Question: Which of the following statement is true for Power-of-2 allocators? <ol> <li>Buddy System splits and recombines memory blocks in a predetermined manner during allocation and deallocation.</li> <li>No splitting of blocks takes place, also no effort is made to coalesce adjoining blocks to form larger blocks; when released, a block is simply returned to its free list.</li> <li> When a request is made for m bytes, the allocator first check the free list containing blocks whose size is 2i for the smallest value of i such that 2i ≥ m.if the free list is empty ,it checks the list containing blocks that are higher the next higher power of 2 in size on so on. an entire block is allocated to a request.</li> <li> When a request is made for m bytes. the system finds the smallest power of 2 that is ≥ m. Let this be 2i.if the list is empty, it checks the lists for block of size 2i+1.it takes one block off this list and splits it into two halves of size 2i.it put one of these blocks into the free list of size 2i,and uses the other block to satisfy the request.</li> </ol>

Answer:

No splitting of blocks takes place, also no effort is made to coalesce adjoining blocks to form larger blocks; when released, a block is simply returned to its free list.
When a request is made for m bytes ,the allocator first check the free list containing blocks whose size is 2ⁱ for the smallest value of i such that 2ⁱ ≥ m. If the free list is empty, it checks the list containing blocks that are higher the next higher power of 2 in size on so on. an entire block is allocated to a request.

MCQ: Which of the following statement is true for Power-of-2 allocators? <ol> <li>Buddy System splits and recombines memory blocks in a predetermined manner during allocation and deallocation.</li> <li>No splitting of blocks takes place, also no effort is made to coalesce adjoining blocks to form larger blocks; when released, a block is simply returned to its free list.</li> <li> When a request is made for m bytes, the allocator first check the free list containing blocks whose size is 2i for the smallest value of i such that 2i ≥ m.if the free list is empty ,it checks the list containing blocks that are higher the next higher power of 2 in size on so on. an entire block is allocated to a request.</li> <li> When a request is made for m bytes. the system finds the smallest power of 2 that is ≥ m. Let this be 2i.if the list is empty, it checks the lists for block of size 2i+1.it takes one block off this list and splits it into two halves of size 2i.it put one of these blocks into the free list of size 2i,and uses the other block to satisfy the request.</li> </ol>

A. 1 only
B. 1,2 only
C. 2,3 only
D. 1,4 only

Correct Answer: A. 1 only

Explanation:

No splitting of blocks takes place, also no effort is made to coalesce adjoining blocks to form larger blocks; when released, a block is simply returned to its free list.
When a request is made for m bytes ,the allocator first check the free list containing blocks whose size is 2ⁱ for the smallest value of i such that 2ⁱ ≥ m. If the free list is empty, it checks the list containing blocks that are higher the next higher power of 2 in size on so on. an entire block is allocated to a request.